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            <span class="title-hover-animation">Lc1621.大小为K的不重叠线段的数目</span>
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            <h2 id="1621-大小为-K-的不重叠线段的数目"><a href="#1621-大小为-K-的不重叠线段的数目" class="headerlink" title="1621. 大小为 K 的不重叠线段的数目"></a><a class="link" target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/number-of-sets-of-k-non-overlapping-line-segments/">1621. 大小为 K 的不重叠线段的数目<i class="fas fa-external-link-alt"></i></a></h2><p>Difficulty: <strong>中等</strong></p>
<p>给你一维空间的 <code>n</code> 个点，其中第 <code>i</code> 个点（编号从 <code>0</code> 到 <code>n-1</code>）位于 <code>x = i</code> 处，请你找到 <strong>恰好</strong> <code>k</code> <strong>个不重叠</strong> 线段且每个线段至少覆盖两个点的方案数。线段的两个端点必须都是 <strong>整数坐标</strong> 。这 <code>k</code> 个线段不需要全部覆盖全部 <code>n</code> 个点，且它们的端点可以重合。<br>请你返回 <code>k</code> 个不重叠线段的方案数。由于答案可能很大，请将结果对 <code>1e9+7</code> <strong>取余</strong> 后返回。</p>
<p><strong>示例 1：</strong></p>
<p><img lazyload="" src="/images/loading.svg" data-src="https://i.loli.net/2020/11/24/CO2b9MBa5WFTNnU.png"></p>
<figure class="highlight c"><table><tr><td class="code"><pre><span class="line">输入：n = <span class="number">4</span>, k = <span class="number">2</span></span><br><span class="line">输出：<span class="number">5</span></span><br><span class="line">解释：</span><br><span class="line">如图所示，两个线段分别用红色和蓝色标出。</span><br><span class="line">上图展示了 <span class="number">5</span> 种不同的方案 {(<span class="number">0</span>,<span class="number">2</span>),(<span class="number">2</span>,<span class="number">3</span>)}，{(<span class="number">0</span>,<span class="number">1</span>),(<span class="number">1</span>,<span class="number">3</span>)}，{(<span class="number">0</span>,<span class="number">1</span>),(<span class="number">2</span>,<span class="number">3</span>)}，{(<span class="number">1</span>,<span class="number">2</span>),(<span class="number">2</span>,<span class="number">3</span>)}，{(<span class="number">0</span>,<span class="number">1</span>),(<span class="number">1</span>,<span class="number">2</span>)} 。</span><br></pre></td></tr></table></figure>

<p><strong>示例 2：</strong></p>
<figure class="highlight c"><table><tr><td class="code"><pre><span class="line">输入：n = <span class="number">3</span>, k = <span class="number">1</span></span><br><span class="line">输出：<span class="number">3</span></span><br><span class="line">解释：总共有 <span class="number">3</span> 种不同的方案 {(<span class="number">0</span>,<span class="number">1</span>)}, {(<span class="number">0</span>,<span class="number">2</span>)}, {(<span class="number">1</span>,<span class="number">2</span>)} 。</span><br></pre></td></tr></table></figure>

<p><strong>示例 3：</strong></p>
<figure class="highlight c"><table><tr><td class="code"><pre><span class="line">输入：n = <span class="number">30</span>, k = <span class="number">7</span></span><br><span class="line">输出：<span class="number">796297179</span></span><br><span class="line">解释：画 <span class="number">7</span> 条线段的总方案数为 <span class="number">3796297200</span> 种。将这个数对 <span class="number">109</span> + <span class="number">7</span> 取余得到 <span class="number">796297179</span> 。</span><br></pre></td></tr></table></figure>

<p><strong>示例 4：</strong></p>
<figure class="highlight c"><table><tr><td class="code"><pre><span class="line">输入：n = <span class="number">5</span>, k = <span class="number">3</span></span><br><span class="line">输出：<span class="number">7</span></span><br></pre></td></tr></table></figure>

<p><strong>示例 5：</strong></p>
<figure class="highlight c"><table><tr><td class="code"><pre><span class="line">输入：n = <span class="number">3</span>, k = <span class="number">2</span></span><br><span class="line">输出：<span class="number">1</span></span><br></pre></td></tr></table></figure>

<p><strong>提示：</strong></p>
<ul>
<li>  <code>2 &lt;= n &lt;= 1000</code></li>
<li>  <code>1 &lt;= k &lt;= n-1</code></li>
</ul>
<h3 id="解法一"><a href="#解法一" class="headerlink" title="解法一"></a>解法一</h3><p>这题题想了好久，最终还是妥协了，看了题解，发现有暴搜+记忆化的解法，试了下，T了</p>
<figure class="highlight java"><table><tr><td class="code"><pre><span class="line"><span class="keyword">int</span> MOD = (<span class="keyword">int</span>)<span class="number">1e9</span>+<span class="number">7</span>;</span><br><span class="line"></span><br><span class="line"><span class="comment">//写个暴搜试试（T了）</span></span><br><span class="line"><span class="function"><span class="keyword">public</span> <span class="keyword">int</span> <span class="title">numberOfSets2</span><span class="params">(<span class="keyword">int</span> n, <span class="keyword">int</span> k)</span> </span>{</span><br><span class="line">    Long[][] dp = <span class="keyword">new</span> Long[n+<span class="number">1</span>][k+<span class="number">1</span>];</span><br><span class="line">    <span class="keyword">return</span> dfs(n, k, dp);</span><br><span class="line">}</span><br><span class="line"></span><br><span class="line"><span class="comment">//n个点放k条线段的方案数</span></span><br><span class="line"><span class="function"><span class="keyword">public</span> <span class="keyword">int</span> <span class="title">dfs</span><span class="params">(<span class="keyword">int</span> n, <span class="keyword">int</span> k, Long[][] dp)</span> </span>{</span><br><span class="line">    <span class="keyword">if</span> (dp[n][k] != <span class="keyword">null</span>) {</span><br><span class="line">        <span class="keyword">return</span> dp[n][k].intValue();</span><br><span class="line">    }</span><br><span class="line">    <span class="keyword">if</span> (k == <span class="number">0</span>) {</span><br><span class="line">        <span class="keyword">return</span> <span class="number">1</span>;</span><br><span class="line">    }</span><br><span class="line">    <span class="keyword">if</span> (k == <span class="number">1</span>) {</span><br><span class="line">        <span class="keyword">return</span> (n-<span class="number">1</span>)*n/<span class="number">2</span>;</span><br><span class="line">    }</span><br><span class="line">    <span class="keyword">long</span> res = <span class="number">0</span>;</span><br><span class="line">    <span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">1</span>; n-i &gt;= k-<span class="number">2</span>; i++) {</span><br><span class="line">        <span class="comment">//前(i+1)个点放一条线段有i种方案（刨开之前的方案）</span></span><br><span class="line">        res += <span class="number">1l</span> * i * dfs(n-i, k-<span class="number">1</span>, dp);</span><br><span class="line">        res = (res + MOD) % MOD;</span><br><span class="line">    }</span><br><span class="line">    dp[n][k] = res;</span><br><span class="line">    <span class="keyword">return</span> (<span class="keyword">int</span>)res;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>

<h3 id="解法二"><a href="#解法二" class="headerlink" title="解法二"></a>解法二</h3><p>N^3dp，<code>dp[i][j]</code>代表长度为i的节点放置j条线段有几种方案，状态的转移分两种情况</p>
<ol>
<li>最后一条线段不覆盖右端点，那么递推就直接<code>dp[i][j] = dp[i-1][j]</code></li>
<li>最后一条线段覆盖右端点，枚举最后一条线段的长度所对应的<code>dp[i-len][j-1]</code>累加起来</li>
</ol>
<figure class="highlight java"><table><tr><td class="code"><pre><span class="line"><span class="function"><span class="keyword">public</span> <span class="keyword">int</span> <span class="title">numberOfSets</span><span class="params">(<span class="keyword">int</span> n, <span class="keyword">int</span> k)</span> </span>{</span><br><span class="line">    <span class="keyword">int</span> MOD = (<span class="keyword">int</span>)<span class="number">1e9</span>+<span class="number">7</span>;</span><br><span class="line">    <span class="keyword">long</span>[][] dp = <span class="keyword">new</span> <span class="keyword">long</span>[n+<span class="number">1</span>][k+<span class="number">1</span>];</span><br><span class="line">    <span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">0</span>; i &lt;= n; i++) {</span><br><span class="line">        dp[i][<span class="number">0</span>] = <span class="number">1</span>;</span><br><span class="line">    }</span><br><span class="line">    <span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">2</span>; i &lt;= n; i++) {</span><br><span class="line">        <span class="keyword">for</span> (<span class="keyword">int</span> j = <span class="number">1</span>; j &lt;= Math.min(i-<span class="number">1</span>, k); j++) {</span><br><span class="line">            <span class="comment">//j没有覆盖右端点的情况</span></span><br><span class="line">            dp[i][j] = dp[i-<span class="number">1</span>][j];</span><br><span class="line">            <span class="comment">//枚举j覆盖区间右端点的所有情况，注意从1开始</span></span><br><span class="line">            <span class="keyword">for</span> (<span class="keyword">int</span> s = <span class="number">1</span>; s &lt; i; s++) {</span><br><span class="line">                dp[i][j] = (dp[i][j] + dp[s][j-<span class="number">1</span>]) % MOD;</span><br><span class="line">            }</span><br><span class="line">        }</span><br><span class="line">    }</span><br><span class="line">    <span class="keyword">return</span> (<span class="keyword">int</span>) dp[n][k];</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<p>这里数据范围是1e3，显然N^3是过不了的，所以需要优化，三重循环种求的其实是前面已经计算过的累计的前缀和，所以可以通过前缀和优化成N^2</p>
<figure class="highlight java"><table><tr><td class="code"><pre><span class="line"><span class="function"><span class="keyword">public</span> <span class="keyword">int</span> <span class="title">numberOfSets</span><span class="params">(<span class="keyword">int</span> n, <span class="keyword">int</span> k)</span> </span>{</span><br><span class="line">    <span class="keyword">int</span> MOD = (<span class="keyword">int</span>)<span class="number">1e9</span>+<span class="number">7</span>;</span><br><span class="line">    <span class="keyword">long</span>[][] dp = <span class="keyword">new</span> <span class="keyword">long</span>[n+<span class="number">1</span>][k+<span class="number">1</span>];</span><br><span class="line">    <span class="keyword">long</span>[][] sum = <span class="keyword">new</span> <span class="keyword">long</span>[n+<span class="number">1</span>][k+<span class="number">1</span>];</span><br><span class="line">    <span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">1</span>; i &lt;= n; i++) {</span><br><span class="line">        dp[i][<span class="number">0</span>] = <span class="number">1</span>;</span><br><span class="line">        sum[i][<span class="number">0</span>] = sum[i-<span class="number">1</span>][<span class="number">0</span>] + <span class="number">1</span>; </span><br><span class="line">    }</span><br><span class="line">    <span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">2</span>; i &lt;= n; i++) {</span><br><span class="line">        <span class="keyword">for</span> (<span class="keyword">int</span> j = <span class="number">1</span>; j &lt;= Math.min(i-<span class="number">1</span>, k); j++) {</span><br><span class="line">            <span class="comment">//j没有覆盖右端点的情况</span></span><br><span class="line">            dp[i][j] = dp[i-<span class="number">1</span>][j];</span><br><span class="line">            <span class="comment">//枚举j覆盖区间右端点的所有情况，注意从1开始</span></span><br><span class="line">            <span class="comment">// for (int s = 1; s &lt; i; s++) {</span></span><br><span class="line">            <span class="comment">//     dp[i][j] = (dp[i][j] + dp[s][j-1]) % MOD;</span></span><br><span class="line">            <span class="comment">// }</span></span><br><span class="line">            dp[i][j] = (dp[i][j] + sum[i-<span class="number">1</span>][j-<span class="number">1</span>]) % MOD;</span><br><span class="line">            sum[i][j] = sum[i-<span class="number">1</span>][j] + dp[i][j];</span><br><span class="line">        }</span><br><span class="line">    }</span><br><span class="line">    <span class="keyword">return</span> (<span class="keyword">int</span>) dp[n][k];</span><br><span class="line">}</span><br></pre></td></tr></table></figure>

<h3 id="解法三"><a href="#解法三" class="headerlink" title="解法三"></a>解法三</h3><p>这种DP的方式更好理解</p>
<ul>
<li><code>dp[i][j][0]</code>: 前i个点放j条线段，第j条线段没有覆盖右端点，所以显然<code>dp[i][j][0] = dp[i-1][j][0] + dp[i-1][j][1]</code></li>
<li><code>dp[i][j][1]</code>: 前i个点放j条线段，第j条线段覆盖右端点（可以延长），首先j位置单独成一条线段（<code>dp[i-1][j-1][0] + dp[i-1][j-1][1]</code>），然后就是j位置和前面的连成一条线段的（<code>dp[i-1][j]</code>）</li>
</ul>
<figure class="highlight java"><table><tr><td class="code"><pre><span class="line"><span class="function"><span class="keyword">public</span> <span class="keyword">int</span> <span class="title">numberOfSets</span><span class="params">(<span class="keyword">int</span> n, <span class="keyword">int</span> k)</span> </span>{</span><br><span class="line">    <span class="keyword">int</span> MOD = (<span class="keyword">int</span>)<span class="number">1e9</span>+<span class="number">7</span>;</span><br><span class="line">    <span class="comment">//dp[i][j][0]: 前i个点放j条线段，第j条线段没有覆盖右端点</span></span><br><span class="line">    <span class="comment">//dp[i][j][1]: 前i个点放j条线段，第j条线段覆盖右端点（可以延长）</span></span><br><span class="line">    <span class="keyword">long</span>[][][] dp = <span class="keyword">new</span> <span class="keyword">long</span>[n+<span class="number">1</span>][k+<span class="number">1</span>][<span class="number">2</span>];</span><br><span class="line">    <span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">0</span>; i &lt;= n; i++) {</span><br><span class="line">        dp[i][<span class="number">0</span>][<span class="number">0</span>] = <span class="number">1</span>;</span><br><span class="line">    }</span><br><span class="line">    <span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">2</span>; i &lt;= n; i++) {</span><br><span class="line">        <span class="keyword">for</span> (<span class="keyword">int</span> j = <span class="number">1</span>; j &lt;= Math.min(i-<span class="number">1</span>, k); j++) {</span><br><span class="line">            dp[i][j][<span class="number">0</span>] = dp[i-<span class="number">1</span>][j][<span class="number">0</span>] + dp[i-<span class="number">1</span>][j][<span class="number">1</span>];</span><br><span class="line">            dp[i][j][<span class="number">1</span>] = dp[i-<span class="number">1</span>][j-<span class="number">1</span>][<span class="number">0</span>] + dp[i-<span class="number">1</span>][j-<span class="number">1</span>][<span class="number">1</span>] + dp[i-<span class="number">1</span>][j][<span class="number">1</span>];</span><br><span class="line">            dp[i][j][<span class="number">0</span>] = (dp[i][j][<span class="number">0</span>] + MOD) % MOD;</span><br><span class="line">            dp[i][j][<span class="number">1</span>] = (dp[i][j][<span class="number">1</span>] + MOD) % MOD;</span><br><span class="line">        }</span><br><span class="line">    }</span><br><span class="line">    <span class="keyword">return</span> (<span class="keyword">int</span>)(dp[n][k][<span class="number">0</span>] + dp[n][k][<span class="number">1</span>] + MOD) % MOD;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<h3 id="解法四"><a href="#解法四" class="headerlink" title="解法四"></a>解法四</h3><p>其实是上面解法的优化递推，省去了前缀和的空间，直接从状态转移方程上进行优化<br><img lazyload="" src="/images/loading.svg" data-src="http://static.imlgw.top/blog/20201126/RxHR7Ushje0p.png?imageslim" alt="mark"><br>列出递推中的连续两项，然后进行合并优化，直接得到递推公式</p>
<figure class="highlight java"><table><tr><td class="code"><pre><span class="line"><span class="comment">//有一点技巧性的递推</span></span><br><span class="line"><span class="function"><span class="keyword">public</span> <span class="keyword">int</span> <span class="title">numberOfSets</span><span class="params">(<span class="keyword">int</span> n, <span class="keyword">int</span> k)</span> </span>{</span><br><span class="line">    <span class="keyword">int</span> MOD = (<span class="keyword">int</span>)<span class="number">1e9</span>+<span class="number">7</span>;</span><br><span class="line">    <span class="comment">//n个点，放k个线段，不重叠的方案数</span></span><br><span class="line">    <span class="keyword">long</span>[][] dp = <span class="keyword">new</span> <span class="keyword">long</span>[n+<span class="number">1</span>][k+<span class="number">1</span>];</span><br><span class="line">    <span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">0</span>; i &lt;= n; i++) {</span><br><span class="line">        dp[i][<span class="number">0</span>] = <span class="number">1</span>;</span><br><span class="line">    }</span><br><span class="line">    </span><br><span class="line">    <span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">2</span>; i &lt;= n; i++) {</span><br><span class="line">        <span class="keyword">for</span> (<span class="keyword">int</span> j = <span class="number">1</span>; j &lt;= i-<span class="number">1</span> &amp;&amp; j &lt;= k; j++) {</span><br><span class="line">            <span class="comment">//dp[2][1] = 2*dp[1][1] - dp[0][1] + dp[1][0]</span></span><br><span class="line">            dp[i][j] = <span class="number">2</span>*dp[i-<span class="number">1</span>][j] - dp[i-<span class="number">2</span>][j] + dp[i-<span class="number">1</span>][j-<span class="number">1</span>];</span><br><span class="line">            dp[i][j] = (dp[i][j] + MOD) % MOD;</span><br><span class="line">        }</span><br><span class="line">    }</span><br><span class="line">    <span class="keyword">return</span> (<span class="keyword">int</span>)dp[n][k];</span><br><span class="line">}</span><br></pre></td></tr></table></figure>

<h3 id="解法五"><a href="#解法五" class="headerlink" title="解法五"></a>解法五</h3><p>数学法，看不懂，求C(n+k-1, 2k)</p>

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        <ol class="nav"><li class="nav-item nav-level-2"><a class="nav-link" href="#1621-%E5%A4%A7%E5%B0%8F%E4%B8%BA-K-%E7%9A%84%E4%B8%8D%E9%87%8D%E5%8F%A0%E7%BA%BF%E6%AE%B5%E7%9A%84%E6%95%B0%E7%9B%AE"><span class="nav-number">1.</span> <span class="nav-text">1621. 大小为 K 的不重叠线段的数目</span></a><ol class="nav-child"><li class="nav-item nav-level-3"><a class="nav-link" href="#%E8%A7%A3%E6%B3%95%E4%B8%80"><span class="nav-number">1.1.</span> <span class="nav-text">解法一</span></a></li><li class="nav-item nav-level-3"><a class="nav-link" href="#%E8%A7%A3%E6%B3%95%E4%BA%8C"><span class="nav-number">1.2.</span> <span class="nav-text">解法二</span></a></li><li class="nav-item nav-level-3"><a class="nav-link" href="#%E8%A7%A3%E6%B3%95%E4%B8%89"><span class="nav-number">1.3.</span> <span class="nav-text">解法三</span></a></li><li class="nav-item nav-level-3"><a class="nav-link" href="#%E8%A7%A3%E6%B3%95%E5%9B%9B"><span class="nav-number">1.4.</span> <span class="nav-text">解法四</span></a></li><li class="nav-item nav-level-3"><a class="nav-link" href="#%E8%A7%A3%E6%B3%95%E4%BA%94"><span class="nav-number">1.5.</span> <span class="nav-text">解法五</span></a></li></ol></li></ol>
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